STRUCTURE OF Nd-142, Nd-143, Nd-145, Nd-146, Nd-148
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (July 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of contradicting theories of the nuclear force , which could not lead to the correct nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for the discovery of nuclear force and structure by applying the laws of electromagnetism.(See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). STRUCTURE OF NEODYMIUM WITH 32 BLANK POSITIONS Naturally occurring neodymium (Nd) is composed of 5 stable isotopes, 142Nd, 143Nd, 145Nd,146Nd and 148Nd, with 142Nd being the most abundant (27.2% natural abundance), and 2 radioisotopes, 144Nd and 150NdS ). Comparing the structure of Nd-120 with the structure of the Tellurium of 52 protons with high symmetry one might see that the structure of Nd with 60 protons (even number) must have the same structure of high symmetry as that of the Tellurium.(See my STRUCTURE OF Te-120...Te-130 ). However it differs fundamentally from the structure of Tellurium, since here the odd number of extra neutrons give a great spin. For example in the Tellurium the odd number of extra neutrons gives S =+1/2 , while the Nd-143 and the Nd-145 of odd number of extra neutrons give S = -7/2 and S = +7/2 respectively. After a careful analysis of the structure of Nd I discovered that it is due to the reduction of the number N of blank positions. In order to understand these difficulties for the structure of Nd let us analyze carefully the diagram of Nd including the top view of the third horizontal plane. TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS ' (n.)........p58....... n50.......p51....n60 ' ' p53........n42........p16......n16......p44.........n54' ' (n) n47........p25........n6........p6........n26.........p48' ' p37 p45........n25........p5........n5........p26........ n46' ' (n) n55........p41.......n15.......p15.......n43...... .p56' ' n57.......p49.......n52...... p59........(n)' Here you can see the following characteristics of the fundamental shapes formed by the nucleons of the central parallelepiped as The p5n5 and n6p6 create the small horizontal square of Mg-24 for creating the central parallelepiped of the alpha particle nuclei. The n15p15 and p16n16 create the first small horizontal rectangle. The p25n25 and p26n26 create the second small horizontal rectangle. The p41, n42, n43 and p44 make the great horizontal square of the great central parallelepiped. The p45, n46, n47 and p48 form the first great horizontal rectangle. The p49, n50, p51 and n52 form the second great horizontal rectangle. Then these great horizontal rectangles which give the structure of Tellurium with 52 protons are able to form 8 blank positions for receiving 8 vertical pn systems. Here you see the p53, n54, n55, p56 n57, p58, p59 and n60 of such vertical pn systems. Also here you see the p37 with 2(n) , while the p38 with 2(n) are not shown, because they are at the fourth horizontal plane. That is, the third and the fourth horizontal plane under the contributions of p37n37 and p38n38 increase the blank positions. Here the p37n37 and p38n38 with opposite spins are moved from the two squares of the diagram in order to give not the 4(n) but 8(n) of weak horizontal bonds existing outside of the central parallelepiped. Note that the bonds of n in other horizontal planes are characterized by a combination of weak and strong bonds existing along the spin axis. Whereas the blank positions of {n} give three bonds per neutron. In the structure of the Tellurium the n54, n55, n57 and n60 of the third plane along with the symmetrical positions of the fourth horizontal plane were able to receive 8 extra neutrons . However in the structure of Nd looking at the top view of the third horizontal plane one concludes that there exist only 4 blank positions of the third and the fourth plane for receiving 4 extra neutrons . Under this condition the p37n37 and p38n38 as vertical pn systems are moved from the squares to the central parallelepiped in order to make more symmetrical vertical rectangles with then47p47 and n48p48 respectively. Thus using the diagram of Nd and the top view of the first, the second, and the third horizontal plane in this new arrangement the number N of blank positions is given by The p39n39 and p40n40 give 6n of strong bonds with opposite spins. The first and the sixth plane give 4(n) of weak bonds with opposite spins. The second and the fifth plane gives 4{n} + 8n of opposite spins. The third and the fourth plane give 4(n) of opposite spins The p37 and p38 give at the third and the fourth plane extra 4(n) That is N = 4{n} + 14n + 12(n) = 30 blank positions. ' ' STRUCTURE OF Nd-142, Nd-146 AND Nd-148 WITH S =0 Since in these cases the extra neutrons have the same number of positive and negative spins we conclude that the S=0 is due to the S=0 of the 60 protons and the 60 neutrons as well as of the S=0 of the extra neutrons of opposite spins. For example the Nd-142 of 22 extra neutrons has 4{n} + 14n + 4(n) of opposite spins, while the Nd-148 of 28 extra neutrons has 4{n} + 14n + 10(n) of opposite spins . ' ' STRUCTURE OF Nd-143 AND Nd-145 In the structure of Nd-143 the p37n37 is moved to make a vertical rectangle with the 545n45. That is, both the p37 and p38 along with 4(n) exist at the fourth plane with negative spins . Especially these 6 nucleons give S =- 6/2 . Under these arrangements we get 32 blank positions able to receive 19 extra neutrons of negative spins and 13 extra neutrons of positive spins. Since the Nd-143 of 23 extra neutrons has S =-7/2 we conclude that it has 15 extra neutrons of negative spins and 8 extra neutrons of positive spins . In the case of Nd-145 the p38n38 is moved from the square to the central parallelepiped to make a symmetrical vertical rectangle with the n46p46. That is, both the p37 and p38 along with 4(n) exist at the third plane with positive spins. Especially these 6 nucleons give S = +6/2. Under these new arrangements we get 32 blank positions able to receive 19 extra neutrons of positive spins and 13 extra neutrons of negative spins . Since the Nd-145 of 25 extra neutrons has S =+7/2 we conclude that it has 16 extra neutrons of positive spins and 9 extra neutrons of negative spins. ' DIAGRAM OF Nd FORMING 32 BLANK POSITIONS' Here the additional p60n60 is not shown, but you can see the position of n60 by using the top view of the third horizontal plane. Here you see the p47n47 along with the p48n48, which make two symmetrical alpha particles of opposite spins . But you cannot see the additional p49n49 the n52p52 of the third alpha particle and the n50p50 and the p51n51 of the fourth alpha particle. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. In the same way the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. ' n40.......p40 .............n n40p40 with n' ' n31………p12.........n12.......p32.......n' ' n........p31........n11.........p11…… n32 Sixth h. plane' ' p29.........n10.........p10…… n30' ' n29……...p9..........n9 …….p30 Fifth h. plane' ' p47.......n27.........p8..........n8.........p28...........n48' ' n45...........p27........n7.........p7........n28..........p46...........(n) Fourth h. plane ' ' n47..........p25.........n6.........p6..........n26...........p48' ' (n)....p45..........n25…….p5..........n5……….p26.........n46 Third h. plane' ' n23………p4...........n4………….p24' ' n........p23…….....n3………....p3………..n24 ...............n Second h. plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22] First h. plane' ' n39......p39........n n39p39 with n' TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN 'HERE THE FIRST EXTRA NEUTRON (n ) OF WEAK BONDS MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34. ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22' ' n33.......p33..... (n)' ' TOP VIEW OF THE SECOND HORIZONTAL PLANE' Here the n of strong bonds near the p14 fills the blank position formed by p51 and p14. While the {n} of three bonds per neutron near the p14 fills the blank position formed by p14, p24 and p44. The 2n near p24 and p23 fill the blank positions formed by p24 and p48 as well as by p23 and p45. Moreover the blank position of {n} near p23 is formed by p23, p13 and p41. Finally the blank position of n near the p13 is formed byp13 and p49. That is we have 2{n} +4n and the same situation occurs at the fifth horizontal plane. ' n' ' n14.......p14........{n}' ' n23.......p4.........n4.........p24..........n ' ' n.......p23........n3........p3.........n24' ' {n}...... p13......n13 ' ' n' ndance), and 2radioisotopes, 144Nd and 150Nd Category:Fundamental physics concepts